Puzzle for February 11, 2022 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit non-negative integer.
Scratchpad
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Hint #1
Add A to both sides of eq.5: B + D – A + A = A + C + A which becomes eq.5a) B + D = 2×A + C eq.2 may be written as: eq.2a) A + D = B + C
Hint #2
Subtract the left and right sides of eq.2a from the left and right sides of eq.5a, respectively: B + D – (A + D) = 2×A + C – (B + C) which becomes B + D – A – D = 2×A + C – B – C which becomes B – A = 2×A – B Add A and B to both sides of the above equation: B – A + A + B = 2×A – B + A + B which makes 2×B = 3×A Divide both sides by 2: 2×B ÷ 2 = 3×A ÷ 2 which makes eq.5b) B = 1½×A
Hint #3
Add E to both sides of eq.3: D – E + E = C + E + E which becomes eq.3a) D = C + 2×E
Hint #4
In eq.5a, replace B with 1½×A (from eq.5b), and D with C + 2×E (from eq.3a): 1½×A + C + 2×E = 2×A + C Subtract 1½×A and C from both sides of the equation above: 1½×A + C + 2×E – 1½×A – C = 2×A + C – 1½×A – C which simplifies to 2×E = ½×A Multiply both sides by 2: 2×E × 2 = ½×A × 2 which makes 4×E = A
Hint #5
In eq.5b, substitute (4×E) for A: B = 1½×(4×E) which makes B = 6×E
Hint #6
Substitute 4×E for A, and 6×E for B in eq.4: C + F = 4×E + 6×E + E which becomes C + F = 11×E Subtract C from each side of the equation above: C + F – C = 11×E – C which becomes eq.4a) F = 11×E – C
Hint #7
eq.6 may be written as: A + D + C = B + E + F Substitute B + C for A + D (from eq.2) in the above equation: B + C + C = B + E + F which becomes B + 2×C = B + E + F Subtract B from each side: B + C + C – B = B + E + F – B which becomes eq.6a) 2×C = E + F
Hint #8
Substitute 11×E – C for F (from eq.4a) into eq.6a: 2×C = E + 11×E – C which becomes 2×C = 12×E – C Add C to both sides of the above equation: 2×C + C = 12×E – C + C which makes 3×C = 12×E Divide both sides by 3: 3×C ÷ 3 = 12×E ÷ 3 which makes C = 4×E
Hint #9
Substitute 4×E for C in eq.4a: F = 11×E – 4×E which makes F = 7×E
Hint #10
Substitute 4×E for C in eq.3a: D = 4×E + 2×E which makes D = 6×E
Solution
Substitute 4×E for A and C, 6×E for B and D, and 7×E for F in eq.1: 4×E + 6×E + 4×E + 6×E + E + 7×E = 28 which simplifies to 28×E = 28 Divide both sides of the above equation by 28: 28×E ÷ 28 = 28 ÷ 28 which means E = 1 making A = C = 4×E = 4 × 1 = 4 B = D = 6×E = 6 × 1 = 6 F = 7×E = 7 × 1 = 7 and ABCDEF = 464617